How would an infinitely powerful laser act?

[Tech_Junkie]

They are working on laser propulsion now. I cant find the video, but they have a super high power IR lasers that can propel a tin plate into the air. Not sure of the science, but its basically only the plate and laser.

Found this article but I'm not sure if its the same system above.

Laser Propulsion: Wild Idea May Finally Shine | Space.com

 

[Flaminpyro]

like this

 

[Tech_Junkie]

Thats the one.

Thanks Jeff.

 

[Cyparagon]

Photons do have mass, that is why a black hole is black, light can't escape the gravitational field. Gravity only affects matter (stuff with mass).

It's more like it warps space-time around it and light follows this warped space-time.

But also, "black holes" aren't black.

Also, hawking radiation.

 

[NickD070]

Those are some interesting articles. Thank you for mentioning/posting them. It should still be noted that the actual propulsion comes from the heating and expansion of air inside the cone not from photons propelling the object.

 

[NickD070]

Two more very good points Cyparagon.

 

[Wolfman29]

Guys, guys, guys - you are all missing the point. While photons don't necessarily have a "real" mass, they have an energy. We know from Einstein's equations that an energy has a "proportional" mass equivalent to it. That is, they have momentum but that is because of their energy and NOT their mass (they don't have any).

 

[Ibuprofen200mg]

^and momentum is conserved, the craft blasting lasers out the back is propelled in the opposite direction of the motion of the photon ...like recoil from a rifle, the photons being the bullets in this analogy

 

[Wolfman29]

There ya go

 

[NickD070]

But those last two comments still rely on photons having mass. An object (or paricle)
cannot have momentum without mass because the formula for total momentum is P = mv
where P is momentum, m is mass and v is velocity.

If the speed is light ~299,792,458 m/s so if you multiply that by the mass of a photon (0) and the total momentum of a photon is 0.

So.... that craft is not being "pushed by photons at all. The model is being pushed by expanding air created by rapid heating on the underside of the model (which is explained in the video). This concept is called "lift".

This is the same concept demonstrated by a radiometer.

The fins inside the radiometer rotate when light is shined on them. This is not because of the momentum of the photons, it is because of the heating of air behind the fins that are painted black pushing the fins in a certain direction. It's quite a nice model for demonstrating this principle actually.

 

[Wolfman29]

While yes, that's how a radiometer works (photons have so little momentum, their effects in normal power densities are minimal), they do carry momentum. They must - after all, they can transfer energy by losing energy themselves.

Further, here is the definition of energy, as from Albert Einstein:

E = (Pc)^2 + mc^2.

P, in this case, is the "relativistic" momentum, which is defined as:

P = p/sqrt(1-v^2/c^2). So if we can prove that P does not equal zero, then p (classical momentum) cannot be equal to zero.

Now, we know that photons have no "rest" mass. So let's set m = 0 in Einstein's equation.

E = (Pc)^2 + 0c^2
E = (Pc)^2

We know that photons carry energy, otherwise they wouldn't interact with anything.

So, the momentum of a photon must be

P = sqrt(E)/c, where E is the energy the photon carries. Now, it isn't much - E is relatively small and c is huge. However, it is something, i.e. P =/= 0. Therefore, "small" p cannot be equal to zero either.

Anyway, how could photons heat things up if they didn't "collide" with atoms and give them kinetic energy? After all, temperature is only the average kinetic energy of a large mass of molecules.

 

[Sigurthr]

While yes, that's how a radiometer works (photons have so little momentum, their effects in normal power densities are minimal), they do carry momentum. They must - after all, they can transfer energy by losing energy themselves.

Further, here is the definition of energy, as from Albert Einstein:

E = (Pc)^2 + mc^2.

P, in this case, is the "relativistic" momentum, which is defined as:

P = p/sqrt(1-v^2/c^2). So if we can prove that P does not equal zero, then p (classical momentum) cannot be equal to zero.

Now, we know that photons have no "rest" mass. So let's set m = 0 in Einstein's equation.

E = (Pc)^2 + 0c^2
E = (Pc)^2

We know that photons carry energy, otherwise they wouldn't interact with anything.

So, the momentum of a photon must be

P = sqrt(E)/c, where E is the energy the photon carries. Now, it isn't much - E is relatively small and c is huge. However, it is something, i.e. P =/= 0. Therefore, "small" p cannot be equal to zero either.

Anyway, how could photons heat things up if they didn't "collide" with atoms and give them kinetic energy? After all, temperature is only the average kinetic energy of a large mass of molecules.

Exactly what I was getting at before. They have no REST mass but due to their energy they have an equivalent mass. I've been meaning to post about this every time this thread came up in my inbox but things kept getting in the way. +1 when it lets me.

 

[NickD070]

But those equations you cited simply prove that a photon has no (conventional) mass...

I think word confusion is causing some of the confusion here.
Rest mass and relativistic mass are totally different.

Rest mass is the basically the "weight" of a particle independent from gravity (also called mass).

Relativistic mass is the sum total quantity of energy in a body or system. So relativistic mass is measurement of energy not conventional rest mass. The concept of relativistic mass is totally different from rest mass.

Photons cannot have any "conventional" momentum. The momentum you are referring to is exerted by radiation pressure, which is unrelated to momentum exerted by mass and velocity

In regards to photons heating things up, this isn't done by friction like it is with a particle with mass and velocity striking another particle.
Light from a laser or anything else excites electrons in the atoms of what they are striking. The electrons of that matter are excited by the electromagnetic energy of light. This causes them to resonate more vigorously because of the electromagnetic influence on electrons. Then the atom as a whole makes a quantum transition from "electronically excited" (just the electrons) to "vibrationally excited" this means that the energy then causes the whole atom to move. That atomic vibration is heat.

I admit that the different meanings of the words has caused us to go back and forth arguing the same points. I apologize for not clarifying that I was referring to (actual) mass, and not the "mass equivalent" of the energy contained by a photon.

Anyway, I appreciate the intellectual debate wolfman, and sorry drummer if you thread got a little side tracked.

 

[NickD070]

And sorry Sigurthr, I started typing that message before you posted yours. Yes, you are correct Sigurthr. It's just so easy to think that a relativistic mass is similar to a rest mass because of the names, but they are completely different. :beer:

 

[Benm]

This is the same concept demonstrated by a radiometer.

These things are awful examples to demonstrate something really.

If you are looking at photon mass, these things should spin the opposite way the mostly do - i.e. the silvery side should spin away from the light, since bouncing the photons off imparts more momentum than absorbing them. The dark side of the blades seem to move away from the light though, implying a more complex mechanism that involves local heating of the blades and all.

The general concept should be that light pushes things out of its way, regardless if that light is absorbed or reflected.

 

[Sigurthr]

And sorry Sigurthr, I started typing that message before you posted yours. Yes, you are correct Sigurthr. It's just so easy to think that a relativistic mass is similar to a rest mass because of the names, but they are completely different. :beer:

No problem. =) Cheers mate :beer: